Weak Law of Large Numbers

Question

The given problem is as follows:

Recall that $\log 2 = \int_0^1 1/(x+1) dx$. Hence, by using a uniform $(0,1)$ generator, approximate $\log 2$. Obtain an error of estimation in terms of a large sample 95% confidence interval. If you have access to the statistical package R, write an R function for the estimate and the error of estimation. Obtain your estimate for 10,000 simulations and compare it to the true value.

My answer:

$$\int_0^1 1/(x+1) dx = (1-0)\int_0^1 1/(x+1) dx/(1-0) = \int_0^1 1/(x+1) f(x) dx = E(1/(x+1))$$

Where f(x)=1, 0

And then I calculated log 2 from the calculator and got 0.6931471806

From R, I got 0.6920717

So, from the weak law of large numbers, we can see that the sample mean is approaching the actual mean as n gets larger.

My Question:

Is my answer correct? Can I use the calculator to approximate log 2?

Thanks in advance

Answer 1

You are supposed to generate $n = 10000$ random variables in R. You could obtain them by typing

x<-runif(10000)

This generate $x_1, \ldots, x_n$ from a uniform $U \left( 0, 1 \right)$.

To get an estimate of $\log \left( 2 \right)$, you compute the sum $$ \frac{1}{n}\sum_{i = 1}^n \frac{1}{x_i + 1} $$ which converges by the law of large number to $E \left[ \frac{1}{1 + x} \right] = \log \left( 2 \right)$